Answer
$$y= t^3+\sin t+12$$
Work Step by Step
Given$$ \frac{d y}{d t}=3 t^{2}+\cos t, \quad y(0)=12$$
Since :\begin{align*}
y&= \int (3 t^{2}+\cos t) dt\\
&= t^3+\sin t +c
\end{align*}
at $t= 0,\ \ \ y= 12 $, then $c= 12$, hence
$$y= t^3+\sin t+12$$